From its formula, one might expect \(\mathrm{CO}\) to be quite polar, but its dipole moment is actually small \((0.11 \mathrm{D})\). (a) Draw the Lewis structure for CO. (b) Calculate the formal charges. (c) Based on your answers to parts (a) and (b), explain why the dipole moment is so small.

The Lewis structure of a molecule is a diagram that shows how the valence electrons are arranged among the atoms in the molecule. For carbon monoxide (CO), we start by determining the total number of valence electrons. Carbon has four valence electrons, and oxygen has six valence electrons, giving a total of ten valence electrons. We initially draw a single bond between carbon and oxygen. Next, we distribute the remaining valence electrons to satisfy the octet rule, which states that atoms tend to have eight electrons in their valence shell. In the case of CO, we end up with a triple bond between carbon and oxygen (one sigma bond and two pi bonds), and each atom has a lone pair of electrons.

This arrangement ensures that both carbon and oxygen achieve their octet, resulting in a stable structure.

Formal charges help us understand the distribution of electrons in a molecule and identify the best Lewis structure. The formal charge of an atom can be calculated using the formula: \[ \text{Formal Charge} = \text{Number of valence electrons} - (\text{Number of non-bonding electrons} + \text{Number of bonds}) \] Applying this to CO:

• For carbon: Valence electrons = 4, Non-bonding electrons = 2, Number of bonds = 3
  Formal Charge = 4 - (2 + 3) = -1
• For oxygen: Valence electrons = 6, Non-bonding electrons = 2, Number of bonds = 3
  Formal Charge = 6 - (2 + 3) = +1

The result shows that carbon has a formal charge of -1 and oxygen has a formal charge of +1. These formal charges imply that the electrons are not evenly distributed and aid in understanding the behavior and reactivity of the molecule.

Electronegativity is a measure of an atom's ability to attract and hold onto electrons within a chemical bond. Oxygen is more electronegative (EN = 3.44) compared to carbon (EN = 2.55), indicating that oxygen tends to attract electrons more strongly. Given the considerable electronegativity difference, one would expect CO to have a significant dipole moment. However, the actual dipole moment of CO is surprisingly small (0.11 D).

The formal charges calculated earlier reveal essential details about electron distribution. Despite the electronegativity difference, the formal charges cause the electrons to be more evenly distributed across the molecule, reducing the expected polarity. This balanced electron distribution leads to the small dipole moment observed in CO. By examining these aspects, we better understand the electron dynamics in carbon monoxide and why its dipole moment is smaller than anticipated.